imone/ARB huggingface dataset:
Question: Let $T_{1}, T_{2}, T_{3}$ be points on a parabola, and $t_{1}, t_{2}, t_{3}$ the tangents to the parabola at these points. Compute the ratio of the area of triangle $T_{1} T_{2} T_{3}$ to the area of the triangle determined by the tangents.
Answer: Choose a Cartesian system of coordinates such that the equation of the parabola is $y^{2}=4 p x$. The coordinates of the three points are $T_{i}\left(4 p \alpha_{i}^{2}, 4 p \alpha_{i}\right)$, for appropriately chosen $\alpha_{i}, i=1,2,3$. Recall that the equation of the tangent to the parabola at a point $\left(x_{0}, y_{0}\right)$ is $y y_{0}=2 p\left(x+x_{0}\right)$. In our situation the three tangents are given by$$ 2 \alpha_{i} y=x+4 p \alpha_{i}^{2}, \quad i=1,2,3 . $$If $P_{i j}$ is the intersection of $t_{i}$ and $t_{j}$, then its coordinates are $\left(4 p \alpha_{i} \alpha_{j}, 2 p\left(\alpha_{i}+\alpha_{j}\right)\right)$. The area of triangle $T_{1} T_{2} T_{3}$ is given by a Vandermonde determinant: $$ \pm \frac{1}{2}\left|\begin{array}{lll} 4 p \alpha_{1}^{2} & 4 p \alpha_{1} & 1 \\ 4 p \alpha_{2}^{2} & 4 p \alpha_{2} & 1 \\ 4 p \alpha_{3}^{2} & 4 p \alpha_{3} & 1 \end{array}\right|=\pm 8 p^{2}\left|\begin{array}{lll} \alpha_{1}^{2} & \alpha_{1} & 1 \\ \alpha_{2}^{2} & \alpha_{2} & 1 \\ \alpha_{3}^{2} & \alpha_{3} & 1 \end{array}\right|=8 p^{2}\left|\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| . $$The area of the triangle $P_{12} P_{23} P_{31}$ is given by$$ \begin{aligned} & \pm \frac{1}{2}\left|\begin{array}{lll}4 p \alpha_{1} \alpha_{2} & 2 p\left(\alpha_{1}+\alpha_{2}\right) & 1 \\4 p \alpha_{2} \alpha_{3} & 2 p\left(\alpha_{2}+\alpha_{3}\right) & 1 \\4 p \alpha_{3} \alpha_{1} & 2 p\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =\pm 4 p^{2}\left|\begin{array}{ll}\alpha_{1} \alpha_{2}\left(\alpha_{1}+\alpha_{2}\right) & 1 \\\alpha_{2} \alpha_{3}\left(\alpha_{2}+\alpha_{3}\right) & 1 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right|=\pm 4 p^{2}\left|\begin{array}{rr}\left(\alpha_{1}-\alpha_{3}\right) \alpha_{2}\left(\alpha_{1}-\alpha_{3}\right) & 0 \\\left(\alpha_{2}-\alpha_{1}\right) \alpha_{3}\left(\alpha_{2}-\alpha_{1}\right) & 0 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =4 p^{2}\left|\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| \text {. } \end{aligned} $$We conclude that the ratio of the two areas is 2 , regardless of the location of the three points or the shape of the parabola.
Question: For each continuous function $f:[0,1] \rightarrow \mathbb{R}$, we define $I(f)=\int_{0}^{1} x^{2} f(x) d x$ and $J(f)=\int_{0}^{1} x(f(x))^{2} d x$. Find the maximum value of $I(f)-J(f)$ over all such functions $f$.
Answer: We change this into a minimum problem, and then relate the latter to an inequality of the form $x \geq 0$. Completing the square, we see that$$ \left.x(f(x))^{2}-x^{2} f(x)=\sqrt{x} f(x)\right)^{2}-2 \sqrt{x} f(x) \frac{x^{\frac{3}{2}}}{2}=\left(\sqrt{x} f(x)-\frac{x^{\frac{3}{2}}}{2}\right)^{2}-\frac{x^{3}}{4} $$Hence, indeed,$$ J(f)-I(f)=\int_{0}^{1}\left(\sqrt{x} f(x)-\frac{x^{\frac{3}{2}}}{2}\right)^{2} d x-\int_{0}^{1} \frac{x^{3}}{4} d x \geq-\frac{1}{16} $$It follows that $I(f)-J(f) \leq \frac{1}{16}$ for all $f$. The equality holds, for example, for $f:[0,1] \rightarrow \mathbb{R}, f(x)=\frac{x}{2}$. We conclude that$$ \max _{f \in \mathcal{C}^{0}([0,1])}(I(f)-J(f))=\frac{1}{16} . $$
Question: Compute the integral$$ \iint_{D} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{2}}, $$where $D$ is the domain bounded by the circles$$ \begin{array}{ll} x^{2}+y^{2}-2 x=0, & x^{2}+y^{2}-4 x=0, \\ x^{2}+y^{2}-2 y=0, & x^{2}+y^{2}-6 y=0 . \end{array} $$
Answer: The domain $D$ is depicted in Figure 71 . We transform it into the rectangle $D_{1}=$ $\left[\frac{1}{4}, \frac{1}{2}\right] \times\left[\frac{1}{6}, \frac{1}{2}\right]$ by the change of coordinates$$ x=\frac{u}{u^{2}+v^{2}}, \quad y=\frac{v}{u^{2}+v^{2}} . $$The Jacobian is Figure 71$$ J=-\frac{1}{\left(u^{2}+v^{2}\right)^{2}} . $$Therefore,$$ \iint_{D} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{2}}=\iint_{D_{1}} d u d v=\frac{1}{12} . $$
Question: Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$. Find the probability that $\sqrt{2+\sqrt{3}} \leq|v+w|$.
Answer: Because the 1997 roots of the equation are symmetrically distributed in the complex plane, there is no loss of generality to assume that $v=1$. We are required to find the probability that$$ |1+w|^{2}=|(1+\cos \theta)+i \sin \theta|^{2}=2+2 \cos \theta \geq 2+\sqrt{3} . $$This is equivalent to $\cos \theta \geq \frac{1}{2} \sqrt{3}$, or $|\theta| \leq \frac{\pi}{6}$. Because $w \neq 1, \theta$ is of the form $\pm \frac{2 k \pi}{1997} k$, $1 \leq k \leq\left\lfloor\frac{1997}{12}\right\rfloor$. There are $2 \cdot 166=332$ such angles, and hence the probability is $\frac{332}{1996}=\frac{83}{499} \approx 0.166$
Question: The numbers $1,2,3,4,5,6,7$, and 8 are written on the faces of a regular octahedron so that each face contains a different number. Find the probability that no two consecutive numbers are written on faces that share an edge, where 8 and 1 are considered consecutive.
Answer: Consider the dual cube to the octahedron. The vertices $A, B, C, D, E, F, G$, $H$ of this cube are the centers of the faces of the octahedron (here $A B C D$ is a face of the cube and $(A, G),(B, H),(C, E),(D, F)$ are pairs of diagonally opposite vertices). Each assignment of the numbers $1,2,3,4,5,6,7$, and 8 to the faces of the octahedron corresponds to a permutation of $A B C D E F G H$, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. The problem requires us to count octagonal circuits that can be formed by eight of these diagonals.Six of these diagonals are edges of the tetrahedron $A C F H$, six are edges of the tetrahedron $D B E G$, and four are long diagonals, joining opposite vertices of the cube. Notice that each vertex belongs to exactly one long diagonal. It follows that an octagonal circuit must contain either 2 long diagonals separated by 3 tetrahedron edges (Figure 108a), or 4 long diagonals (Figure 108b) alternating with tetrahedron edges.
Figure 108When forming a (skew) octagon with 4 long diagonals, the four tetrahedron edges need to be disjoint; hence two are opposite edges of $A C F H$ and two are opposite edges of $D B E G$. For each of the three ways to choose a pair of opposite edges from the tetrahedron $A C F H$, there are two possible ways to choose a pair of opposite edges from tetrahedron $D B E G$. There are $3 \cdot 2=6$ octagons of this type, and for each of them, a circuit can start at 8 possible vertices and can be traced in two different ways, making a total of $6 \cdot 8 \cdot 2=96$ permutations.An octagon that contains exactly two long diagonals must also contain a three-edge path along the tetrahedron $A C F H$ and a three-edge path along tetrahedron the $D B E G$. A three-edge path along the tetrahedron the $A C F H$ can be chosen in $4 !=24$ ways. The corresponding three-edge path along the tetrahedron $D B E G$ has predetermined initial and terminal vertices; it thus can be chosen in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8 \cdot 24 \cdot 2=384$ permutations of this type.In all, there are $96+384=480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\frac{480}{8 !}=\frac{1}{84}$.
Question: Let $V$ be the vector space of all real $3 \times 3$ matrices and let $A$ be the diagonal matrix

$$
\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1
\end{array}\right) .
$$

Calculate the determinant of the linear transformation $T$ on $V$ defined by $T(X)=\frac{1}{2}(A X+X A)$.


Answer:  Let $X=\left(x_{i j}\right)$ be any element of $M_{3}(\mathbb{R})$. A calculation gives

$$
T(X)=\left(\begin{array}{ccc}
x_{11} & 3 x_{12} / 2 & x_{13} \\n3 x_{21} / 2 & 2 x_{22} & 3 x_{23} / 2 \\nx_{31} & 3 x_{32} / 2 & x_{33}
\end{array}\right) .
$$

It follows that the basis matrices $M_{i j}$ are eigenvectors of $T$. Taking the product of their associated eigenvalues, we get $\operatorname{det} T=2(3 / 2)^{4}=81 / 8$.
Question: An old woman went to the market and a horse stepped on her basket and smashed her eggs. The rider offered to pay for the eggs and asked her how many there were. She did not remember the exact number, but when she had taken them two at a time there was one egg left, and the same happened when she took three, four, five, and six at a time. But when she took them seven at a time, they came out even. What is the smallest number of eggs she could have had? 
Answer: We are to find the smallest positive solution to the system of congruences$$ \begin{aligned} &x \equiv 1(\bmod 60), \\ &x \equiv 0(\bmod 7) . \end{aligned} $$The general solution is $7 b_{1}+420 t$, where $b_{1}$ is the inverse of 7 modulo 60 and $t$ is an integer. Since $b_{1}$ is a solution to the Diophantine equation $7 b_{1}+60 y=1$, we find it using Euclid's algorithm. Here is how to do it: $60=8 \cdot 7+4,7=1 \cdot 4+3,4=1 \cdot 3+1$. Then$$ \begin{aligned} 1 &=4-1 \cdot 3=4-1 \cdot(7-1 \cdot 4)=2 \cdot 4-7=2 \cdot(60-8 \cdot 7)-7 \\ &=2 \cdot 60-17 \cdot 7 \end{aligned} $$Hence $b_{1}=-17$, and the smallest positive number of the form $7 b_{1}+420 t$ is $-7 \cdot 17+$ $420 \cdot 1=301$.